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I am aware of the relationship between the eigenvalues of a matrix and the matrix square (being the eigenvalues squared) but what relationship, if any, is there between the eigenvectors of the matrix and its square? Does it depend on what type of matrix i.e. hermitian, unitary etc?

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    $\begingroup$ Any eigenvector of $A$ is an eigenvector of $A^2$. $\endgroup$ – Lord Shark the Unknown May 26 '17 at 17:05
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If A is a linear transformation from vector space V to itself, with eigenvalue $\lambda$, there exist a non-trivial vector v, such that $Av= \lambda v$. Then $A(Av)= A(\lambda v)$. So $A^2v= \lambda (Av)= \lambda(\lambda v)= \lambda^2 v$. That is if $\lambda$ is an eigenvalue of A then $\lambda^2$ is an eigenvalue of $A^2$.

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  • $\begingroup$ It's worth noting that an eigenvector of $A^2$ doesn't need to be an eigenvector of $A$, keeping in mind nilpotent matrices and generalized eigenvectors. $\endgroup$ – Roland May 26 '17 at 17:23
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By the definition of eigenvector, which is sent by the matrix in a scaled version of itself, you see that doing this twice (or any amount of times you want) still respects the condition. And it is easy to see, as you have pointed out, that the new eigenvalue is the power of the original eigenvalue with as exponent the number of times you apply the matrix.

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